Example of null hypothesis for chi square test

The table below includes the row and column totals. Observed and expected counts are often presented together in a contingency table. In the table below, expected values are presented in parentheses. We can determine the p-value by constructing a chi-square distribution plot with 1 degree of freedom and finding the area to the right of 0. There is not evidence that gender and whether or not an individual has completed an online course are related. Note that we cannot say for sure that these two categorical variables are independent, we can only say that we do not have evidence that they are dependent.

If you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data. After data entry the procedure is the same for both data entry methods. Research question : Is there a relationship between where a student sits in class and whether they have ever cheated? All expected values are at least 5 so we can use the Pearson chi-square test statistic.

There is not evidence of a relationship in the population between seat location and whether a student has cheated. Let's conduct an hypothesis test using the dataset: fallstdata.

Assumption: All expected counts are at least 5. The expected counts here are Our p value is greater than the standard 0. As you may recall, a Chi-square test of independence is method that tests the degree to which one nominal variable is independent of another nominal variable. Example 7. Problem A public opinion poll surveyed a simple random sample of voters. Respondents were classified by gender male or female and by voting preference Republican, Democrat, or Independent. Results are shown in the contingency table below.

Is there a gender gap? Do the men's voting preferences differ significantly from the women's preferences? Use a 0. Here is a template for writing a null-hypothesis for a Chi-square test of independence: Here is a template for writing a null-hypothesis for a Chi-square test of independence: [Variable A] and [Variable B] are independent of one another.

Here is a template for writing a null-hypothesis for a Chi-square test of independence: Gender and voting preferences are independent of one another. Kristin Chavez Nov. I can recommend a site that has helped me. VaibhaviBhosale Apr. Arash Mashhadi Mar. Student at Ayatollah amoli. Fail to reject the null hypothesis meaning the test has not identified a consequential relationship between the two phenomena.

Your email address will not be published. Save my name, email, and website in this browser for the next time I comment. Skip to content. Related posts: Quick Answer: What is the null hypothesis for normality test? Quick Answer: How is Chi square analysis used in genetics?

What is Parametric vs nonparametric? The survey was completed by graduates and the following data were collected on the exercise question:. Based on the data, is there evidence of a shift in the distribution of responses to the exercise question following the implementation of the health promotion campaign on campus?

In this example, we have one sample and a discrete ordinal outcome variable with three response options. We specifically want to compare the distribution of responses in the sample to the distribution reported the previous year i. We now run the test using the five-step approach. The null hypothesis again represents the "no change" or "no difference" situation. If the health promotion campaign has no impact then we expect the distribution of responses to the exercise question to be the same as that measured prior to the implementation of the program.

Notice that the research hypothesis is written in words rather than in symbols. The research hypothesis as stated captures any difference in the distribution of responses from that specified in the null hypothesis.

We do not specify a specific alternative distribution, instead we are testing whether the sample data "fit" the distribution in H 0 or not. We must first assess whether the sample size is adequate.

Specifically, we need to check min np 0 , np 1, Thus, min 0. The sample size is more than adequate so the formula can be used. The appropriate critical value is 5. We now compute the expected frequencies using the sample size and the proportions specified in the null hypothesis. We then substitute the sample data observed frequencies and the expected frequencies into the formula for the test statistic identified in Step 2.

The computations can be organized as follows. Notice that the expected frequencies are taken to one decimal place and that the sum of the observed frequencies is equal to the sum of the expected frequencies. The test statistic is computed as follows:. We reject H 0 because 8. Here, we reject H 0 and concluded that the distribution of responses to the exercise question following the implementation of the health promotion campaign was not the same as the distribution prior.

The test itself does not provide details of how the distribution has shifted. A comparison of the observed and expected frequencies will provide some insight into the shift when the null hypothesis is rejected.

Does it appear that the health promotion campaign was effective? If the null hypothesis were true i. Thus, there is a shift toward more regular exercise following the implementation of the health promotion campaign.

There is evidence of a statistical difference, is this a meaningful difference? Is there room for improvement? The distribution was based on specific values of body mass index BMI computed as weight in kilograms over height in meters squared.

Suppose we want to assess whether the distribution of BMI is different in the Framingham Offspring sample. We must assess whether the sample size is adequate. The sample size is more than adequate, so the formula can be used. The appropriate critical value is 7. We then substitute the sample data observed frequencies into the formula for the test statistic identified in Step 2. We organize the computations in the following table.

We reject H 0 because Here we show that the distribution of BMI in the Framingham Offspring Study is different from the national distribution. To understand the nature of the difference we can compare observed and expected frequencies or observed and expected proportions or percentages. The frequencies are large because of the large sample size, the observed percentages of patients in the Framingham sample are as follows: 0.

Are these meaningful differences? In the module on hypothesis testing for means and proportions, we discussed hypothesis testing applications with a dichotomous outcome variable in a single population. We presented a test using a test statistic Z to test whether an observed sample proportion differed significantly from a historical or external comparator. The chi-square goodness-of-fit test can also be used with a dichotomous outcome and the results are mathematically equivalent.

In the prior module, we considered the following example. Here we show the equivalence to the chi-square goodness-of-fit test.

An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?